1. Prove it for n=1. Im sure u can do this yourself
2. Assume it is true for n=k. Again u can state this i am sure. You should have something like:
k^3+2k=12m where m is some integer
3. Prove it true for n=k+2. This is because n is even (as in the question) and assuming for n=k+1 would mean n is odd. I'm assuming this is where you're having problems? Ok then. You''re aiming to prove
(k+2)^3+2(k+2)= 12xsomething
LHS= (k+2)^3+2(k+2)
= k^3 + 6k^2 +12k +8 +2k +4
=12m + 6k^2+ 12k + 12
now 6=3x2 therefore 6k^2=3(2xk^2) as k^2 is even therefore 2xeven= something divsible by 4 (this should be clear). As such 2k^2=4j where j is random integer
Therefore
=12m +12j +12k +12
=12(m+j+k+1)
= RHS
4. Just state the standard proof statement.